On a dry road, a car with good tires may be able to brake with a constant deceleration of 3.31 m/s2. (a) How long does such a car, initially traveling at 28.2 m/s, take to stop? (b) How far does it travel in this time?
0 = 28.2 - 3.31 t
solve for t
then use t
d = 28.2 t - (1/2)(3.31) t^2
To find the answers to these questions, we can use the equations of motion, specifically the equation for constant acceleration:
v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement
(a) To find how long it takes for the car to stop, we need to determine the final velocity when it comes to rest. The final velocity is 0 m/s, and the initial velocity is given as 28.2 m/s. The acceleration of the car is -3.31 m/s^2 (negative because it is deceleration).
Using the equation of motion, we can rearrange it to solve for time (t):
0^2 = 28.2^2 + 2(-3.31)s
Rearranging the equation further:
784.56 = 2(-3.31)s
-6.62s = 784.56
Dividing both sides by -6.62:
s = 784.56 / (-6.62)
s ≈ -118.61 m/s^2
Since displacement cannot be negative, we take the positive value for s, which is approximately 118.61 m/s^2.
Now, we can substitute the value of s into the equation again to find time (t):
0 = 28.2 + 2(-3.31)(118.61)
0 = 28.2 - 2(392.9926)
0 = 28.2 - 785.9852
28.2 = 785.9852
Simplifying further:
t = 785.9852 / 28.2
t ≈ 27.85 seconds
Therefore, the car takes approximately 27.85 seconds to stop.
(b) To determine how far the car travels in this time, we can use the equation of motion:
s = ut + 0.5at^2
Substituting the given values:
s = 28.2(27.85) + 0.5(-3.31)(27.85)^2
Simplifying:
s = 782.67 - 0.5(27.85)(921.3225)
s = 782.67 - 12.95(921.3225)
s = 782.67 - 11928.481
s ≈ -11145.811
Since distance cannot be negative, we take the positive value for s, which is approximately 11145.811 meters.
Therefore, the car travels approximately 11145.811 meters (or about 11.15 kilometers) during this time.